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Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other.There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - \(C^6_4*(C^2_1)^4\)

look at other problem:

Quote:

Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team?

we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team.

Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

total ways of choosing 4 out of 12:12x11x10x9first person can be chosen in 12 ways. The next person can be chosen in 10 ways (because we don't want spouses to be in the group). The next person can be again chosen in only 8 ways (out of the 10 people left, we have to exclude 2 whose spouses we have already selected). And the last person in 6 ways.

Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

= (12C1 * 10C1 * 8C1*6C1)/4!/ 12C4 = 12*10*8*6 / (12*11*10*9) = 16/33

suresh, can you please explain the logic of this calculation? Thank you!

12C1 = select any person from 6 married couples (12 person)10C1 = select second person from remaining people and exclude the first person's spouse8C1 = select 3rd person from remain people exclue first and second perssons's spouses6C1 = select 4th person from remaining people exclude 1st,2nd ,3rd persons's spouse

Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other.There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - \(C^6_4*(C^2_1)^4\)

look at other problem:

Quote:

Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team?

we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team.

I hope it is clearer now.

Wow, makes very clear sense now. Much obliged.Ali

another way of looking at this problem :

total number of ways = 12C4 = 495

now let's try to find out total number of unfavourable ways. We can subtract those from 495 to get the number of favourable ways.

number of ways of choosing 4 people such that there are two couples = 6C2 = 15

number of ways of choosing 4 people such that there is only one couple = number of ways of choosing one couple * number of ways of choosing two people who are not couples = 6C1 (10C2 -5) = 240.

In x2suresh's solution, I'm unable to visualize the exact arrangements because of which we should divide 12x10x8x6 ways by 4!. Could someone please explain in more detail ? The thing is, we're choosing here, so how does the question of arrangements come in picture at all ?